san=a 1( 1-p^n)/( 1-p),sbn=b 1( 1-q^n)/( 1-q)
From Cn=an+bn, sn = San+sbn.
=a 1( 1-p^n)/( 1-p)+b 1( 1-q^n)/( 1-q)
sn/s(n- 1)=[a 1( 1-q)+b 1( 1-q)-a 1( 1-q)p^n-b 1( 1-p)q^n]/
[a 1( 1-q)+b 1( 1-q)-a 1( 1-q)p^(n- 1)-b 1( 1-p)q^(n- 1]]
Sequence {an}{bn} is a geometric series composed of positive numbers, so p>q>0 can be divided into two categories according to the relationship between P and 1.
1 hour1>: p > Q, hour, n-> oo, p n-> 0,q^n—>； 0,p^(n- 1)—>； 0,q^(n- 1)—>； 0, so:
Serial number (n-1)-> [a1(1-q)+b1(1-q)]/[a1-]
2. When p> 1, 1/p
sn/s(n- 1)=[a 1( 1-q)( 1/p)^(n- 1)+b 1( 1-q)( 1/p)^(n- 1)-a 1( 1-q)p-b 1( 1-p)q(q/p)^(n- 1)]/
【a 1( 1-q)( 1/p)^(n- 1)+b 1( 1-q)( 1/p)^(n- 1)-a 1( 1-q)p-b 1( 1-p)(q/p)^(n- 1))]—>； p
To sum up: when P
When p> is at 1, limsn/s (n- 1) = p,
(2) Solution: (1) Proof: When n >; At 1
3t*Sn-(2t+3)S(n- 1)=3t
3t*S(n+ 1)-(2t+3)Sn=3t
Subtract two expressions: 3t*a(n+ 1)-(2t+3)*an=0.
Therefore, a (n+1)/an = 2/3+1/t (constant).
In addition, 3t * S2-(2t+3) s1= 3t * (a2+a1)-(2t+3) * a1= 3t.
Arrangement: A2/A 1 = 2/3+ 1/t (constant)
Therefore, {an} is a geometric series.
(2) F (t) = 2/3+ 1/t, so BN = 2/3+B (n- 1).
Bn-b(n- 1)=2/3 (constant) {bn} is a arithmetic progression, the first term is 1, and the tolerance is 2/3.
So: bn =1+2 (n-1)/3 = (1+2n)/3.
(3) Because the tolerance of bn is 2/3, and b(k+2)-b(k)=4/3, {B2N} is a arithmetic progression with b2=5/3 as the first term and the tolerance is 4/3.
The original formula = b1b2-b2b3+b3b4b5-...+b2n-1b2n-b2nb2n+1.
= B2(b 1-B3)+B4(B3-b5)+…+b2n(b2n- 1-b2n+ 1)
=(-4/3)(b2+b4+…+b2n)
=(-4/3){ 5n/3+[n *(n- 1)/2]* 4/3 }
=-(8n^2+ 12n)/9